Django CMS菜单，如何将类仅设置为父类? [英] Django CMS menu, how to set a class to parents only?

问题描述

我的菜单需要一些工作.我需要在父菜单项上有一个类，但由于递归，它无法正常工作.

My menu needs some work. I need to have a class on the parent menu items, but with my recursion, it is not working.

这是我的菜单的一部分:

This is (part of) my menu:

Home
    Teachers (id teachers)
        Contact
        Info
        Projects
          myproject
          yourproject

我从这样的老师"开始:

I start with "teachers" like this:

{% show_menu_below_id "teqchers" 0 1 0 1 "teachers_menu.html" %}

这是我的Teachers_menu.html:

And this is my teachers_menu.html:

{% load menu_tags %}
{% for child in children %} 
    <li class="{% if child.selected %}selected parent_{{forloop.counter}}{% endif %} {% if child.sibling %}parent_{{forloop.counter}} {% endif %}"> 
        <a href="{{ child.get_absolute_url }}">{{ child.get_menu_title }}</a>
        {% if child.children %}
        <div class="submenu">   
            <ul>
                {% show_menu from_level to_level extra_inactive extra_active template "" "" child %}
            </ul>
        </div>
        {% endif %}
    </li>
{% endfor %}

有了这个，我的菜单就工作了一段时间.

With this in place, my menu is working for a bit.

当我单击项目"时，一切都很好，并且可以看到2个项目.但是，当我单击一个项目时，我希望页面显示出来，但是没有，它会重新构建菜单，并将所需的类添加到子元素中:

When i click Projects, all is well and the 2 projects are in view. But when i click a project, i expect the page to show, but it does not, it does rebuild my menu, and adds the needed class to the child elements:

{% if child.selected %}selected parent_{{forloop.counter}}{% endif %}

很明显，因为我想现在是个孩子，但是如何防止这种情况呢?我只需要第一个菜单项的那个类.

Obvious because it is a child now i guess, but how to prevent this? I only need that class for the first menu items.

推荐答案

而不是这样做:

{% show_menu from_level to_level extra_inactive extra_active template "" "" child %}

我现在添加了另一个这样的模板:

I now added another template like this:

{% show_menu from_level to_level extra_inactive extra_active "teachers_submenu.html" "" "" child %}

在该模板中，我现在有:

And in that template I now have:

{% load menu_tags %}
{% for child in children %} 
    <li class="{% if child.selected %}selected{% endif %}"> 
        <a href="{{ child.get_absolute_url }}">{{ child.get_menu_title }}</a>
    </li>
{% endfor %}

因此，额外的模板将处理子菜单.经过一些样式设置，现在可以使用.

So the extra template takes care of the submenu. With some styling it now works.

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